I think the answer is C because the bar in class 2 is shorter than the bar in class 1, I might be wrong though.
ANSWER:
Let t = logtan[x/2]
⇒dt = 1/ tan[x/2] * sec² x/2 × ½ dx
⇒dt = 1/2 cos² x/2 × cot x/2dx
⇒dt = 1/2 * 1/ cos² x/2 × cosx/2 / sin x/2 dx
⇒dt = 1/2 cosx/2 / sin x/2 dx
⇒dt = 1/sinxdx
⇒dt = cosecxdx
Putting it in the integration we get,
∫cosecx / log tan(x/2)dx
= ∫dt/t
= log∣t∣+c
= log∣logtan x/2∣+c where t = logtan x/2
At the beginning, the chances that the teacher will choose a student who has forgotten his lunch are 2/10, or 1/5.
If the teacher now chooses another student from the remaining 9 students, the chances of his choosing a student who has forgotten his lunch are 1/9, since one of the forgetful students has already been removed from the original group of 10 students, and there are only 9 in the group remaining.
The probability that both students forgot lunch is (1/5)(1/9), or 1/45.
It would be B. "Regular" factoring without the quadratic formula can't be done. Completing the square will work easily since the coefficient on the squared term is a 1. I personally wouldn't graph this to find the solutions, although you could with and x/y chart to give you enough coordinates to plot the graph correctly.
Answer: If 
and 
, then 
Step-by-step explanation:
Let be "a", "b", and "c" all real numbers, the Transitive property of equalities states that :
If and , then
You can observe that if two numbers are equal to the same number, then all this numbers are equal to each other.
For example, applying the Transitive property of equalities to:
and 
You can equate them:
And solve for "x":
